Answer
a. $x=-2$ and $x=1$.
b. The multiplicity for zero $x=-2$ is $1$ and is odd.
The multiplicity for zero $x=1$ is $1$ and is odd.
c. The maximum possible number of turning points is $1$.
d. See graph
Work Step by Step
a.
Factoring out $\frac{1}{3}$ of the polynomial:
$$f(x)=\frac{1}{3}(x^2-x-2)$$
Set $f(x)=0$:
$$0=\frac{1}{2}(x^2+x-2)$$ $$0=x^2+x-2$$ $$0=(x+2)(x-1)$$ $$x+2=0$$ $$x=-2$$ $$x-1=0$$ $$x=1$$ Thus, the real zeros are $x=-2$ and $x=1$.
b.
$$f(x)=\frac{1}{2}(x+2)^1(x-1)^1$$ From the power of the factor $(x+2)^1$ which is $1$, the multiplicity for zero $x=-2$ is $1$ and is odd.
From the power of the factor $(x-1)^1$ which is $1$, the multiplicity for zero $x=1$ is $1$ and is odd.
c.
Since the degree of the polynomial $\frac{1}{3}x^2+\frac{1}{3}x-\frac{2}{3}$ is $2$, the maximum possible number of turning points is: $$n-1=2-1=1$$
d. The graph of the function is as shown and it shows the zeros are $x=-2$ and $x=1$, the multiplicity of zero $x=-2$ is odd since the graph crosses the $x$-axis and the multiplicity of zero $x=1$ is odd since the graph crosses also the $x$-axis, and the number of turning points is $1$.
